MO Bundesrunde Klasse 9 Mock Paper 4: Questions, Hints and Solutions

Put S=a+b+c. Every one of a,b,c divides S. Pairwise coprimality implies their product divides S, so abc≤S. If a≥2, then b,c≥2. For fixed b,c≥2 the difference abc−a−b−c increases with a, and the same holds for each variable separately, because bc−1>0. At (2,2,2) that difference is 8−6=2>0. Hence abc>S whenever all three are at least 2, a contradiction. Thus a=1. The condition c divides b+1 gives c≤b+1, and the ordering gives c≥b. If c=b, coprimality forces b=c=1. If c=b+1, the condition b divides c+1=b+2 forces b to divide 2, so b=1 or 2. This yields (1,1,2) and (1,2,3). All three triples, including (1,1,1), are pairwise coprime and satisfy the divisibilities.

Conclusion: Exactly $(1,1,1),(1,1,2),(1,2,3)$.

Review the idea: Greatest common divisor · Divisibility

Testing (1,1,1) gives 3+6t≥0, so t≥−1/2. Testing (1,−1,0) gives 2−2t≥0, so t≤1. Conversely, expand the identity $$x^2+y^2+z^2+2t(xy+yz+zx)=\frac{1-t}{3}\bigl((x-y)^2+(y-z)^2+(z-x)^2\bigr)+\frac{1+2t}{3}(x+y+z)^2.$$ Both coefficients are nonnegative exactly for −1/2≤t≤1, proving sufficiency. For −1/2<t<1 both are positive, so equality forces x=y=z and x+y+z=0, hence x=y=z=0. At t=−1/2 only the first coefficient remains, so equality means x=y=z. At t=1 only the second remains, so equality means x+y+z=0. These conditions also visibly suffice in the displayed identity.

Conclusion: Exactly −1/2≤t≤1, with equality as described by the two endpoint forms.

Review the idea: Sum of squares · Inequality rules and signs

Use coordinates relative to vectors AB and AC: the pair (x,y) denotes the point reached from A by adding x·AB and y·AC. These coordinates are well defined because the triangle is nondegenerate. A point (x,y) has perpendicular height above AB equal to y times C’s height, and height above AC equal to x times B’s height; this follows by taking perpendicular components of the two side vectors. Thus for an interior point T=(x,y), $[ABT]/[ABC]=y$ and $[ACT]/[ABC]=x$. In these coordinates A=(0,0), B=(1,0), C=(0,1). The given ratios yield D=(1/3,2/3), E=(0,1/3), F=(2/3,0). The three cevian equations are $$\begin{gathered}AD:y=2x,\\ BE:x+3y=1,\\ CF:3x+2y=2.\end{gathered}$$ Solving in pairs gives U=(1/7,2/7), V=(4/7,1/7), W=(2/7,4/7). They are distinct and interior. The height formulas give $[ABU]/[ABC]=2/7$, $[ACW]/[ABC]=2/7$, and $[BCV]/[ABC]=1-[ABV]/[ABC]-[ACV]/[ABC]=1-1/7-4/7=2/7$. These three outer triangles and triangle UVW partition ABC, as their boundaries follow the three cevians. Subtracting their areas gives $[UVW]/[ABC]=1-6/7=1/7$.

Conclusion: The enclosed triangle has one seventh of the area of ABC.

Review the idea: Triangle area ratios · Concurrency and collinearity

Choose P with the largest number d of wins, and let S be the set of the d players P beat. Suppose some other player Q fails the required property. Then P did not beat Q, so Q beat P. Moreover no player in S beat Q; otherwise P would reach Q in two wins. Since every pair has a winner, Q therefore beat every player in S as well. Those d players and P are all distinct, so Q has at least d+1 wins, contradicting maximality of d. Thus P has the desired property. When there is one player, the statement is vacuous and that player works.

Conclusion: A player with the maximum number of wins has the required property.

Review the idea: Pigeonhole principle · Proof methods

Swap coordinates if needed so a≤b, and put $k=(a^2+b^2+1)/(ab)$, a positive integer. The integer $b^{\prime}=ka-b=(a^2+1)/b$ is positive. The quadratic equation $b^2-kab+a^2+1=0$ shows, by substituting its other root b′, that $a^2+(b^{\prime})^2+1=kab^{\prime}$; hence the new pair has the same k. If a=1, b divides 2, so b=1 or 2 and k=3. If a≥2, equality b=a is impossible because it would require $a^{2}$ to divide 1. Thus b≥a+1, giving $$0\lt b^{\prime}=\frac{a^2+1}{b}\le\frac{a^2+1}{a+1}=a-1+\frac2{a+1}\lt a.$$ Replacing (a,b) by (b′,a) strictly decreases its largest entry. Repeating must reach a pair with smaller entry 1, proving k=3 for the original pair. Reversing each step uses $(u,v)\mapsto(v,3v-u)$. Starting at (1,1) gives (1,2),(2,5),(5,13),…; it preserves the equation and positive ordered entries. Every sorted solution descends to (1,1) or (1,2), and the latter is the first generated pair. Thus this rule, together with swaps, generates all solutions.

Conclusion: The quotient is always 3; generate ordered pairs by (u,v)↦(v,3v−u) from (1,1), and allow swaps.

Review the idea: Factorisation integer solutions · Strong induction

If P is the constant c, the equation says c=$c^{2}$, giving P=0 or P=1. Otherwise let its degree be d≥1 and leading coefficient a≠0. The degrees of the two sides are $d^{2}$ and 2d, so d=2. Their leading coefficients are then $a^{3}$ and $a^{2}$, forcing a=1. Write $P(x)=x^2+bx+c$. The original equation becomes $$P(x)^2+bP(x)+c=P(x)^2,$$ hence bP(x)+c is the zero polynomial. Since P is nonconstant, its quadratic coefficient here forces b=0, and then c=0. We obtain P(x)=$x^{2}$. Substitution verifies all three candidates, so the list is complete.

Conclusion: Exactly P=0, P=1 and P(x)=$x^{2}$.

Review the idea: Polynomial functions · Functions inverses and composition